Questions and Answers #1 Laplace Transform

Questions and Answers Sheet 1 Laplace Transform Question #1 Determine the inverse Laplace transform of the function below. 3 (2𝑠+7)4 3 𝐿−1 {(2𝑠+7)4} = ??? Answer: 3 𝐿−1 {(2𝑠+7)4} = 𝐿−1 { = = = = = 3 24 3 24 3 24 3 24 1 𝐿−1 { 7 4 2 3 7 4 2 24 (𝑠+ ) } } (𝑠+ ) 1 3! 3! (𝑠+ ) × 𝐿−1 { × × × 1 3! 1 3! 3 7 3+1 2 × 𝐿−1 { 7 3! 7 3+1 (𝑠+ ) 2 } } × 𝑒 2𝑡 × 𝑡 3 16×3×2×1 3 −1 7 × 𝑒 2𝑡 × 𝑡 3 𝐿 {(2𝑠+7)4} = 1 32 7 × 𝑒 2𝑡 × 𝑡 3 Question #2 Use the accompanying tables of Laplace transform and properties of Laplace transform of the function below. 𝑡𝑒 𝑡 cos 6 𝑡 Answer: Question #3 Find the inverse Laplace transform of the function: 𝐹(𝑠) = 2𝑠 3 +9𝑠 2 +11𝑠+3 (𝑠+1)3 𝑠 Answer: Step 1 Given: 𝐹(𝑠) = 2𝑠 3 +9𝑠 2 +11𝑠+3 (𝑠+1)3 𝑠 Step 2 To determine: Inverse Laplace transform of the given function:=? Step 3 Method used: We use the partial fraction method to solve the given function; First of all, We find the partial fraction of the given function; Then use some formulas of inverse Laplace transformations: 1 𝐿−1 { } = 𝑒 −𝑡 𝑠+1 2 𝐿−1 { } = 2𝑡𝑒 −𝑡 (𝑠 + 1)2 𝑒 −𝑡 𝑡 2 1 𝐿−1 {(𝑠+1)3} = 3 𝐿−1 { } 𝑠 2 = 3𝐻(𝑡) Step 4 Solution to the question: Take the partial fraction of: 2𝑠 3 +9𝑠 2 +11𝑠+3 =− (𝑠+1)3 𝑠 1 2 𝑠+1 1 + (𝑠+1)2 + (𝑠+1)3 + 3 𝑠 Now: Taking Laplace inverse of the resultant function: 1 2 1 3 𝐿−1 {− + (𝑠+1)2 + (𝑠+1)3 + } 𝑠+1 𝑠 Step 5 Now: Solving this expression: 1 2 1 3 𝐿−1 {− + (𝑠+1)2 + (𝑠+1)3 + } = 𝑠+1 −1 1 −𝐿 { } 𝑠+1 𝑠 1 −1 1 1 + 2𝐿 {(𝑠+1)2} + 𝐿−1−1 {(𝑠+1)3} + 3𝐿−1 { } 𝑠 Substituting all the values as: 1 𝐿−1 { } = 𝑒 −𝑡 𝑠+1 2 𝐿−1 {(𝑠+1)2} = 2𝑡𝑒 −𝑡 𝑒 −𝑡 𝑡 2 1 𝐿−1 {(𝑠+1)3} = 3 𝐿−1 { } 𝑠 2 = 3𝐻(𝑡) Hence, We get: 𝐿−1 (𝐹(𝑠)) = = −𝑒 −𝑡 2𝑠 3 +9𝑠 2 +11𝑠+3 + 2𝑡𝑒 −𝑡 (𝑠+1)3 𝑠 𝑒 −𝑡 𝑡 2 + 2 + 3𝐻(𝑡) So, This is the Laplace inverse of the given function; Question #4 Let F(s) be the Laplace transform of 𝑓(𝑥) = 𝑡 cos(2𝑡) find the value of F(1) Answer: Step 1 Determine the Laplace transform of 𝑓(𝑡) = 𝑡 cos(2𝑡) 𝐿{𝑓(𝑡)} = 𝐿{𝑡 cos(2𝑡)} 𝐹(𝑠) = − =− 𝑑 ( 𝑠 𝑑𝑠 𝑠 2 +22 (𝑠 2 +4)−2𝑠 2 (𝑠 2 +4)2 𝑠 2 −4 = (𝑠2 +4)2 ) Step 2 𝑠 2 −4 Substitute 1 for s in the equation 𝐹(𝑠) = (𝑠2 +4)2 12 − 4 𝐹(1) = 2 (1 + 4)2 1−4 = 2 5 −3 = 25 −3 Therefore 𝐹(1) = 25 Question #5 Find the inverse Laplace transform 𝑓(𝑡) = 𝐿−1 {𝐹(𝑠)} of the function 𝐹(𝑠) = 8𝑒 −4𝑠 𝑠 2 +64 Use h(t-c) for the Heaviside function ℎ𝑐 (𝑡) if necessary. 𝑓(𝑡) = 𝐿−1 { 8𝑒 −4𝑠 𝑠 2 +64 }= Answer: Step 1 Consider the provided function, 𝐹(𝑠) = 8𝑒 −4𝑠 𝑠 2 +64 The inverse laplace transform of F(s) is find as, 𝑓(𝑡) = 𝐿−1 (𝐹(𝑠)) = 𝐿−1 ( 8𝑒 −4𝑠 𝑠 2 +64 ) Apply inverse transform rule, if 𝐿−1 (𝐹(𝑠)) = 𝑓(𝑡) then 𝐿−1 (𝑒 −𝑐𝑠 𝐹(𝑠)) = ℎ(𝑡 − 𝑐)𝑓(𝑡 − 𝑐) where ℎ𝑐 (𝑡) is Heaviside function. For 8𝑒 −4𝑠 𝑠 2 +64 ; 𝐹(𝑠) = 8 𝑠 2 +64 ,𝑐 = 4 Step 2 Use the inverse Laplace transform table, 𝐿−1 ( So, 𝑎 ) = sin(𝑎𝑡) 𝑠 2 +𝑎2 8 𝐿−1 ( 2 ) 𝑠 +64 = 𝐿−1 ( Thus, 𝑓(𝑡) = 𝐿−1 ( 8 𝑠 2 +82 8𝑒 −4𝑠 𝑠 2 +64 ) = sin(8𝑡) ) = ℎ(𝑡 − 4) sin(8(𝑡 − 4)) Question #6 To find the Laplace transform of 𝑒 9𝑡 sin ℎ (𝑎𝑡) you would substitute __ into Answer: Step 1 Calculate the Laplace transform of 𝑒 9𝑡 sin ℎ (𝑎𝑡) 𝐿{𝑒 9𝑡 sin ℎ (𝑎𝑡)} = [𝐿{sin ℎ (𝑎𝑡)}]𝑠⇒𝑠−9 Step 2 Since 𝐿{sin ℎ (𝑎𝑡)} = 𝑎 𝐿{𝑒 9𝑡 sin ℎ (𝑎𝑡)} = [ ] 𝑠 2 −𝑎2 𝑎 , therefore 𝑠 2 −𝑎2 𝑠⇒𝑠−9 𝑎 𝑠 2 −𝑎2 which gives ____ 𝑎 = (𝑠−9)2 −𝑎2 Hence, you would substitute s=s-9 into 𝑎 𝑠 2 −𝑎2 which gives 𝑎 (𝑠−9)2 −𝑎2 Question #7 Using Laplace Transform, evaluate the integral-differential equation 𝑦 ′′ (𝑥) + 9𝑦(𝑥) = 40𝑒 𝑥 ; 𝑦(0) = 5, 𝑦 ′ (0) = −2 Answer: Step 1 Solve the given differential equation by Laplace transform. The given equation is 𝑦 ′′ (𝑥) + 9𝑦(𝑥) = 40𝑒 𝑥 ⋯ (1) The given equation is 𝑦(0) = 5 and 𝑦 ′ (0) = −2 Step 2 Taking the Laplace transform on both sides of given equation. 𝐿{𝑦 ′′ (𝑥) + 9𝑦(𝑥)} = 𝐿{40𝑒 𝑥 } 𝐿{𝑦 ′′ (𝑥)} + 9𝐿{𝑦(𝑥)} = 40𝐿(𝑒 𝑥 ) 1 1 {𝑠 2𝑦 − 𝑠𝑦(0) − 𝑦 ′ (0)} + 9𝑦 = 40{ } {𝐿(𝑒 𝑎𝑥 ) = } 𝑠 2𝑦 − 𝑠(5) − (−2) + 9𝑦 = 40 𝑠 2𝑦 − 5𝑠 + 2 + 9𝑦 = 𝑦(𝑠 2 + 9) = 𝑦(𝑠 2 + 9) = 40 𝑠−1 𝑠−1 𝑠−𝑎 {𝑦(0) = 5 and 𝑦 ′ (0) = −2} 𝑠−1 + 5𝑠 − 2 𝑠−1 40+(5𝑠−2)(𝑠−1) 𝑠−1 5𝑠 2 −7𝑠+42 𝑦 = (𝑠−1)(𝑠2 40 +9) ⋯ (2) Step 3 Solve the R.H.S. of equation (2) by partial fraction. 5𝑠 2 −7𝑠+42 𝐴 = + 𝐵𝑠+𝐶 (𝑠−1)(𝑠 2 +9) 𝑠−1 𝑠 2 +9 2 𝐴(𝑠 +9)+(𝐵𝑠+𝐶)(𝑠−1) = = = (𝑠−1)(𝑠 2 +9) 𝐴𝑠 2 +9𝐴+𝐵𝑠 2 −𝐵𝑠+𝐶𝑠−𝐶 (𝑠−1)(𝑠 2 +9) (𝐴+𝐵)𝑠 2 +(−𝐵+𝐶)𝑠+(9𝐴−𝐶) (𝑠−1)(𝑠 2 +9) Equating the coefficients of same variables. A+B=5 -B+C=-7 9A-C=42 Hence, a=5 , c=-6 , b=1 5𝑠 2 −7𝑠+42 = (𝑠−1)(𝑠 2 +9) 4 𝑠−1 + 𝑠−6 𝑠 2 +9 ⋯ (3) Step 4 Equating equation (2) and (3). 4 𝑠−6 𝑦= + 2 { from (2) and (3)} 𝑠−1 𝑠 +9 Taking the inverse transform 4 𝑠−6 𝐿−1 {𝑦} = 𝐿−1 { + 2 } 𝑦 = 4𝐿−1 { 1 𝑠−1 𝑥 𝑠−1 𝑠 +9 𝑠 } + 𝐿−1 { 𝑠 2 +32 1 } − 6𝐿−1 { 1 𝑠 2 +32 } = 4𝑒 + cos 3 𝑥 − 6 ⋅ sin 3 𝑥 3 = 4𝑒 𝑥 + cos 3 𝑥 − 2 sin 3 𝑥 Hence, the solution of the equation is 4𝑒 𝑥 + cos 3 𝑥 − 2 sin 3 𝑥 Question #8 Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace transform of the function below. (1 + 𝑒 −4𝑡 )2 Answer: We have to calculate the laplace transform of (1 + 𝑒 −4𝑡 )2 Now (1 + 𝑒 −4𝑡 )2 = (1)2 + (𝑒 −4𝑡 )2 + 2 ⋅ 1 ⋅ 𝑒 −4𝑡 = 1 + 𝑒 −8𝑡 + 2𝑒 −4𝑡 now 𝐿{(1 + 𝑒 −4𝑡 )2 } = 𝐿{1 + 𝑒 −8𝑡 + 2𝑒 −4𝑡 } = 𝐿{1} + 𝐿{𝑒 −8𝑡 } + 𝐿{2𝑒 −4𝑡 } = 𝐿{1} + 𝐿{𝑒 −8𝑡 } + 2𝐿{𝑒 −4𝑡 } 1 1 1 1 = + +2 {as 𝐿{𝑒 −𝑎𝑡 } = } 𝑝 𝑝+8 𝑝+4 𝑝+𝑎 1 1 2 ∴ 𝐿{(1 + 𝑒 −4𝑡 )2 } = + + 𝑝 𝑝+8 𝑝+4 Question #9 Use Laplace transforms to solve the following initial value problem. 𝑥 ′′ + 𝑥 = 3 cos 3 𝑡, 𝑥(0) = 1, 𝑥 ′ (0) = 0 The solution is x(t)=? Answer: Step 1 The given differential equation is 𝑥 ′′ + 𝑥 = 3 cos(3𝑡) Initial condition: x(0)=1 and x'(0)=0 Solve the differential equation by using laplace transform. Step 2 Consider 𝑥 ′′ + 𝑥 = 3 cos(3𝑡) Operate laplace operator on the both sides, then 𝐿{𝑥 ′′ (𝑡)} + 𝐿{𝑥(𝑡)} = 𝐿{3 cos(3𝑡)} 3𝑠 𝑠 2 𝑥(𝑠) − 𝑠𝑥 ′ (0) − 𝑥 ′ (0) + 𝑥(𝑠) = 2 𝑠 2 𝑥(𝑠) + 𝑥(𝑠) − 1 = (𝑠 2 + 1)𝑥(𝑠) = 3𝑠 𝑠 2 +9 𝑠 2 +3𝑠+9 𝑥(𝑠) = (𝑠2 𝑠 +9 3𝑠 𝑠 2 +9 +1 +9)(𝑠 2 +1) Step 3 By using partial fraction method, 𝑠 2 + 3𝑠 + 9 𝑥(𝑠) = 2 (𝑠 + 9)(𝑠 2 + 1) 3 𝑠 8 +9) = ((𝑠2 3 𝑠 8 +1) ) + ((𝑠2 1 ) + (𝑠2 +1) Operate laplace inverse on the both sides, then 3 𝑠 8 +9) 𝐿−1 {𝑥(𝑠)} = 𝐿−1 ((𝑠2 3 3 𝑠 8 +1) ) + 𝐿−1 ((𝑠2 3 = cos(3𝑡) + cos(𝑡) + sin(𝑡) 8 3 8 3 ⇒ 𝑥(𝑡) = cos(3𝑡) + cos(𝑡) + sin(𝑡) 8 8 1 ) + 𝐿−1 ((𝑠2 ) +1) Hence,the required solution is 3 3 𝑥(𝑡) = cos(3𝑡) + cos(𝑡) + sin(𝑡) 8 8 Question #10 Consider the following initial value problem: 𝑦 ′′ − 4𝑦 ′ − 45𝑦 = sin(7𝑡) , 𝑦(0) = 1, 𝑦 ′ (0) = −4 Using Y for the Laplace transform of y(t), i.e., 𝑌 = 𝐿𝑦(𝑡), find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s)=? Answer: Step 1 The given initial value problem is 𝑦 ′′ − 4𝑦 ′ − 45𝑦 = sin(7𝑡) , 𝑦(0) = 1, 𝑦 ′ (0) = −4 Apply Laplace transform on both sides of the given equation as follows. 𝐿{𝑦 ′′ } − 4𝐿{𝑦 ′ } − 45𝐿{𝑦} = 𝐿{sin 7 𝑡} 7 𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) − 4(𝑠𝑌(𝑠) − 𝑦(0)) − 45𝑌(𝑠) = 2 𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0) − 4𝑠𝑌(𝑠) + 4𝑦(0) − 45𝑌(𝑠) = 2 𝑠 𝑌(𝑠) − 𝑠(1) − (−4) − 4𝑠𝑌(𝑠) + 4(1) − 45𝑌(𝑠) = 7 2 𝑠 𝑌(𝑠) − 𝑠 + 4 − 4𝑠𝑌(𝑠) + 4 − 45𝑌(𝑠) = 7 𝑠 +49 7 𝑠 2 +49 𝑠 2 +49 𝑠 2 +49 Step 2 On further simplification, 𝑠 2 𝑌(𝑠) − 𝑠 + 8 − 4𝑠𝑌(𝑠) − 45𝑌(𝑠) = (𝑠 2 − 4𝑠 − 45)𝑌(𝑠) + 8 − 𝑠 = (𝑠 2 − 4𝑠 − 45)𝑌(𝑠) = (𝑠 2 − 4𝑠 − 45)𝑌(𝑠) = 7 𝑠 2 +49 − (8 − 𝑠) 𝑠 2 +49 𝑠 3 −8𝑠 2 +49𝑠−385 𝑠 3 −8𝑠 2 +49𝑠−385 𝑌(𝑠) = (𝑠2 7 7 𝑠 2 +49 𝑠 2 +49 −4𝑠−45)(𝑠 2 +49) Question #11 Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace transform of the function below. 𝑒 −2𝑡 cos 6 𝑡 + 𝑒 5𝑡 − 1 Answer: Step 1 𝐿{𝑒 −2𝑡 cos 6 𝑡 + 𝑒 5𝑡 − 1} 1 𝐿{𝑒 𝑎𝑡 } = 𝑠−𝑎 𝑠 𝐿{cos 𝑏 𝑡} = 2 𝑠 + 𝑏2 𝑠−𝑎 𝑎𝑡 𝐿{𝑒 cos 𝑏 𝑡} = (𝑠−𝑎)2 2 +𝑏 𝑠+2 1 1 = + − (𝑠 + 2)2 + 62 𝑠 − 5 𝑠 𝑠+2 𝐿{𝑒 −2𝑡 cos 6 𝑡 + 𝑒 5𝑡 − 1} = (𝑠+2)2 +36 + 1 𝑠−5 − 1 𝑠 Question #12 Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below. 6𝑡 4 𝑒 −6𝑡 − 𝑡 4 + cos 6 𝑡 Answer: 𝐿{6𝑡 4 𝑒 −6𝑡 − 𝑡 4 + cos 6 𝑡} 𝐿{𝑡 𝑛 } = 𝑛! 𝑠 𝑛+1 1 𝐿{𝑒 𝑎𝑡 } = 𝑠−𝑎 𝐿(cos 6 𝑡) = 4! 𝑠 𝑠 2 +36 4! = 6 ((𝑠+6)4+1 ) − 𝑠 4+1 + 𝑠 𝑠 2 +36 𝐿{6𝑡 4 𝑒 −6𝑡 − 𝑡 4 + cos 6 𝑡} = 144 = (𝑠+6)5 − 24 + 𝑠5 𝑠 𝑠 2 +36 Question #13 Using (1) or (2), find L(f) if f(t) equals: sin2 𝜔 Answer: find Laplace transformation: Concept: Formulas: 1 1) 𝐿[1] = , 𝑠 > 0 𝑠 𝑠 2) 𝐿[cos 𝑎 𝑡] = 𝑠 2 +𝑎2 ,𝑠 > 0 We have to find 𝐿[sin2 𝜔𝑡] ∴ we know that cos 2𝑡 = 1 − 2 sin2 ∴ 2 sin2 𝑡 = 1 − cos 2 𝑡 1 sin2 𝑡 = (1 − cos 2 𝑡) 2 1 ∴ sin2 𝑡 = (1 − cos 2 𝑡) 2 2 1 ∴ sin 𝜔 𝑡 = (1 − cos 2 𝜔𝑡) 2 1 ∴ 𝐿[sin2 𝜔 𝑡] = 𝐿[1 − cos 2 𝜔𝑡] 2 1 = [𝐿[1] − 𝐿[cos 2 𝜔𝑡]] 2 1 1 𝑠 2 𝑠 1 1 𝑠 2 +(2𝜔)2 𝑠 2 𝑠 2 𝑠 2 +4𝜔2 1 1 = [ − = [ − ] ] 𝐿[sin 𝜔 𝑡] = ( − 2 𝑠 𝑠 𝑠 2 +4𝜔2 ) Question #14 Find the Inverse Laplace Transform f(t) of the following if F(s) equals: 1 𝑠 4 −4𝑠 2 Answer: Step 1 Solution Given 𝐹(𝑥) = 1 𝑠 4 +4𝑠 2 we need to find the inverse laplace transform Step 2 𝐹(𝑥) = 1 𝑠 4 +4𝑠 2 = 1 𝑠 2 (𝑠 2 +4) it can be written as 𝐴 𝐵 𝐹(𝑥) = 2 + (𝑠2 𝑠 +4) obtain the values of A and B 𝐴= 𝐵= 𝑒 −2𝑡 16 𝑒 2𝑡 16 and 1 1 𝑠 𝑠 2 +4 𝐿−1 (𝐹) = 𝐿−1 ( 2) − 9𝐿−1 ( 𝑡 𝑒 −2𝑡 4 16 𝐿−1 (𝐹) = − − + ) 𝑒 2𝑡 16 Question #15 𝐿−1 ( 5𝑠+2 𝑠 2 +9 )= 1 a) 5 cos 𝑡 + 𝑠𝑖𝑛 3 1 b) 5 cos 9 𝑡 + sin 9 𝑡 3 2 c) 5 cos 3 𝑡 + sin 3 𝑡 3 d) 5 cos 3 𝑡 + 3 sin 2 𝑡 Answer: Step 1 Use the standard formula of inverse laplace transform. Step 2 solution: 𝐿−1 ( = = 5𝑠+2 ) 𝑠 2 +9 5𝑠 2 −1 𝐿 ( 2 ) + 𝐿−1 ( 2 ) 𝑠 +9 𝑠 +9 𝑠 1 5𝐿−1 ( 2 ) + 2𝐿−1 ( 2 ) 𝑠 +9 𝑠 +9 2 = 5 cos 3 𝑡 + sin 3 𝑡 3 So, option (c) is correct Question #16 Represent the following function using unit step functions and find its Laplace transform. 2 x 2 , 0  x  2 f ( x) =   x + 3, x  2 Answer: Step 1 Introduction 1.Piecewise functions are the function, whose function expression is different in different interval of its domain. 2.Function whose value is zero for negative input and 1 for positive input are called Unit step function. Step 2 Computation  f ( x), 0  x  x1 f ( x) =  1 , can be written in unit step function as  f 2 ( x), x1  x Piecewise function 𝑓(𝑥) = 𝑓1 (𝑥) + 𝑢(𝑥 − 𝑥1 )(𝑓2 (𝑥) − 𝑓1 (𝑥)) 2 x 2 , 0  x  2 So, the given function f ( x) =  can be written in unit step function as  x + 3, x  2 𝑓(𝑥) = 2𝑥 2 + 𝑢(𝑥 − 2)(𝑥 + 3 − 2𝑥 2 ) Step 3 Computation Now, to find Laplace transform of given function f(x), solve the equation 𝐿(𝑓) = 𝐿(2𝑥 2 ) + 𝐿[𝑢(𝑥 − 2)(𝑥 + 3 − 2𝑥 2 )] . Use the property 𝐿(𝑢(𝑡 − 𝑎)𝑓(𝑡)) = 𝑒 −𝑠𝑡 ⋅ 𝐿(𝑔(𝑡 + 𝑎)) and 𝐿(𝑡 𝑛 ) = 𝑛! 𝑠 𝑛+1 𝐿(𝑓) = 𝐿(2𝑥 2 ) + 𝐿[𝑢(𝑥 − 2)(𝑥 + 3 − 2𝑥 2 )] = 2𝐿(𝑥 2 ) + 𝐿(𝑢(𝑥 − 2)(𝑥 + 3 − 2𝑥 2 )) 2! ) + 𝑒 −2𝑠 𝐿(𝑥 + 2) + 3 − 2(𝑥 𝑠 2+1 4 + 𝑒 −2𝑠 𝐿(𝑥 + 5 − 2𝑥 2 − 8 − 8𝑥) 𝑠3 4 + 𝑒 −2𝑠 𝐿(−2𝑥 2 − 3 − 7𝑥) 𝑠3 4 2! 3 7 −2𝑠 = 2( = = = = = 𝑠3 4 𝑠3 4 𝑠 +𝑒 +𝑒 (−2 ( −2𝑠 (− 4 𝑠3 𝑠 2+1 3 + 2)2 ) − − 2) 𝑠 𝑠 7 − − 2) 𝑠 𝑠 4 3 7 𝑠 𝑠 𝑠 −2𝑠 ( 3 + + 2) 3 −𝑒 Question #17 Solve the ordinary differential equation with initial conditions using Laplace transform. 𝑦 ′′ + 4𝑦 ′ + 13𝑦 = 20𝑒 −𝑡 , 𝑦0 = 1, 𝑦0′ = 3 The solution of the above equation is given as 𝑦 = 𝐴𝑒 𝐵𝑥 + 𝑒 𝐶𝑥 sin 3 𝑡 − 𝑒 𝐷𝑥 cos 3 𝑡 Answer: Step 1 The given equation is 𝑦 ′′ + 4𝑦 ′ + 13𝑦 = 20𝑒 −𝑡 The initial conditions are y(0)=1 , y'(0)=3 Taking Laplace transform on both sides 𝐿{𝑦 ′′ + 4𝑦 ′ + 13𝑦} = 𝐿{20𝑒 −𝑡 } Apply formula Laplace transform for the derivatives 𝐿{𝑓 ′ (𝑡)} = 𝑠𝐿{𝑓(𝑡)} − 𝑓(0) 𝐿{𝑓 ′′ (𝑡)} = 𝑠 2 𝐿{𝑓(𝑡)} − 𝑠𝑓(0) − 𝑓 ′ (0) 𝐿{𝑒 −𝑡 } = 1 𝑠+1 Step 2 Using the formulas 𝐿{𝑦 ′′ + 4𝑦 ′ + 13𝑦} = 𝐿{20𝑒 −𝑡 } [𝑠 2 𝐿{𝑦} − 𝑠𝑦(0) − 𝑦 ′ (0)] + 4[𝑠𝐿{𝑦} − 𝑦(0)] + 13𝐿{𝑦} = 20𝐿(𝑒 −𝑡 ) 1 [𝑠 2 𝐿{𝑦} − 𝑠(1) − 3] + 4[𝑠𝐿{𝑦} − 1] + 13𝐿{𝑦} = 20 2 𝑠 𝐿{𝑦} − 𝑠(1) − 3 + 4𝑠𝐿{𝑦} − 4 + 13𝐿{𝑦} = 20 𝑠 2 𝐿{𝑦} − 𝑠 + 4𝑠𝐿{𝑦} + 13𝐿{𝑦} − 7 = 20 1 𝑠+1 1 𝑠+1 𝑠+1 𝑠 2 𝐿{𝑦} + 4𝑠𝐿{𝑦} + 13𝐿{𝑦} − 𝑠 − 7 = 20 𝐿{𝑦}{𝑠 2 + 4𝑠 + 13} − 𝑠 − 7 = 20 1 𝑠+1 𝑠+1 Step 3 20 +𝑠+7 𝑠+1 20+𝑠 2 +7𝑠+𝑠+7 ( ) 𝐿{𝑦}{𝑠 2 + 4𝑠 + 13} = 𝐿{𝑦} = 𝐿{𝑦} = 1 𝑠 2 +4𝑠+13 𝑠+1 1 𝑠 2 +8𝑠+27 ( 𝑠 2 +4𝑠+13 𝑠+1 𝑠 2 +8𝑠+27 𝐿{𝑦} = (𝑠+1)(𝑠2 ) +4𝑠+13) Take inverse Laplace 𝑠 2 +8𝑠+27 𝐿−1 {𝐿{𝑦}} = 𝐿−1 {(𝑠+1)(𝑠2 𝑠 2 +8𝑠+27 𝑦 = 𝐿−1 {(𝑠+1)(𝑠2 } +4𝑠+13) } ⋯ (1) +4𝑠+13) Step 4 Making the partial fractions 𝑠 2 + 8𝑠 + 27 𝑎 𝑏𝑠 + 𝑐 = + (𝑠 + 1)(𝑠 2 + 4𝑠 + 13) (𝑠 + 1) (𝑠 2 + 4𝑠 + 13) Multiply both sides by (𝑠 + 1)(𝑠 2 + 4𝑠 + 13) 𝑠 2 + 8𝑠 + 27 𝑎 𝑏𝑠 + 𝑐 = + (𝑠 + 1)(𝑠 2 + 4𝑠 + 13) (𝑠 + 1) (𝑠 2 + 4𝑠 + 13) (𝑠 2 +8𝑠+27)(𝑠+1)(𝑠 2 +4𝑠+13) (𝑠+1)(𝑠 2 +4𝑠+13) = 𝑎(𝑠+1)(𝑠 2 +4𝑠+13) (𝑠+1) + (𝑏𝑠+𝑐)(𝑠+1)(𝑠 2 +4𝑠+13) (𝑠 2 +4𝑠+13) (𝑠 2 + 8𝑠 + 27) = 𝑎(𝑠 2 + 4𝑠 + 13) + (𝑏𝑠 + 𝑐)(𝑠 + 1) (𝑠 2 + 8𝑠 + 27) = 𝑎(𝑠 2 + 4𝑠 + 13) + (𝑏𝑠 2 + 𝑐𝑠 + 𝑏𝑠 + 𝑐) (𝑠 2 + 8𝑠 + 27) = (𝑎 + 𝑏)𝑠 2 + (4𝑎 + 𝑏 + 𝑐)𝑠 + 13𝑎 + 𝑐 Step 5 Equating both sides to get a+b=1 4a+b+c=8 13a+c=27 Solve for a, b and c to get the following values a=2 b=-1 c=1 Therefore, 𝑠 2 +8𝑠+27 (−1)𝑠+1 2 (𝑠+1)(𝑠 2 +4𝑠+13) = (𝑠+1) + (𝑠2 +4𝑠+13) 𝑠 2 +8𝑠+27 −1 Equation (1) reduces to 𝐿 {(𝑠+1)(𝑠2 2 (−1)𝑠+1 } = 𝐿−1 {(𝑠+1) + (𝑠2 +4𝑠+13) } +4𝑠+13) Step 6 (−1)𝑠 + 1 2 2 −𝑠 + 1 𝐿−1 { + 2 } = 𝐿−1 { } + 𝐿−1 { 2 } ⋯ (2) (𝑠 + 1) (𝑠 + 4𝑠 + 13) (𝑠 + 1) (𝑠 + 4𝑠 + 13) −𝑠+1 𝑠−1 Again Rewrite (𝑠2 = − (𝑠2 +4𝑠+13) 3 𝑠+2 = − ((𝑠+2)2 +9) + ((𝑠+2)2 +4𝑠+4+9) +9) Equation (2) reduces to (−1)𝑠+1 2 𝐿−1 {(𝑠+1) + (𝑠2 −1 2 2 𝑠+2 +9 𝑠+2 −1 1 = 𝐿 {(𝑠+1)} + 𝐿 {− (𝑠+2)2 } + 3𝐿 {(𝑠+2)2 } = 2𝑒 −𝑡 −𝑒 −2𝑡 cos(3𝑡) + +9 1 3 𝑒 −2𝑡 3 −2𝑡 = 2𝑒 −𝑡 − 𝑒 −2𝑡 cos(3𝑡) + 𝑒 Step 7 3 } = 𝐿−1 {(𝑠+1)} + 𝐿−1 {− (𝑠+2)2 } + 𝐿−1 {(𝑠+2)2 } +4𝑠+13) −1 +9 sin(3𝑡) sin(3𝑡) +9 comparing with the given equation The answer is: A=2 B=-2 C=-2 D=-2 Question #18 𝐿−1 𝑠−4 4(𝑠 − 3)2 + 16 Answer: Step 1 𝑠−1 4(𝑠−3)2 +16 = = (𝑠−3)−1 4{(𝑠−3)2 +4} (𝑠 − 3) 1 − 2 2 4{(𝑠 − 3) + 2 } 4{(𝑠 − 3)2 + 22 } (𝑠−1) 𝑠−3 } = 𝐿−1 { } 4{(𝑠−3)2 +22 } 4{(𝑠−3)2 +22 } (𝑠−3) 1 −1 1 1 𝐿 {(𝑠−3)2 2} − 𝐿−1 {(𝑠−3)2 2 } 4 +2 4 +2 1 3𝑡 −1 3 1 3𝑡 −1 1 𝑒 𝐿 { 2 2} − 𝑒 𝐿 { 2 2} 4 𝑠 +2 4 𝑠 +2 Now, 𝐿−1 { = = − 𝐿−1 { 1 [\text{ using shifting property}] 1 1 = 𝑒 3𝑡 cos 2 𝑡 − 𝑒 3𝑡 sin 2 𝑡 4 4 𝑠 −1 ∵𝐿 { 2 } = cos 𝑎 𝑡 𝑠 + 𝑎2 1 𝐿−1 { 2 } = sin 𝑎 𝑡 𝑠 + 𝑎2 Step 2 This is the inverse Laplace transformation Question #19 solve the given initial value problem 𝑦 ′′′ + 3𝑦 ′′ − 13𝑦 ′ − 15𝑦 = 0, 𝑦(0) = 𝑦 ′ (0) = 0, 𝑦 ′′ (0) = 1 Answer: 𝑦 ′′′ + 3𝑦 ′′ − 13𝑦 ′ − 15𝑦 = 0, 𝑦(0) = 𝑦 ′ (0) = 0, 𝑦 ′′ (0) = 1 Using Laplace transform in both sides we get 𝑠 3 𝑌(𝑠) − 1 + 3𝑠 2 𝑌(𝑠) − 13𝑠𝑌(𝑠) − 15𝑌(𝑠) = 0 ⇒ (𝑠 3 + 3𝑠 2 − 13 − 15)𝑌(𝑠) − 1 = 0 1 ⇒ 𝑌(𝑠) = 3 2 1 𝑠 +3𝑠 −13𝑠−15 = (𝑠+1)(𝑠−3)(𝑠+5) = −1 1 16 𝑠+1 + 1 1 + 1 1 32 𝑠−3 32 𝑠+5 1 1 1 1 𝑦(𝑡) = − 𝐿−1 { } + 𝐿−1 { } 16 𝑠+1 32 𝑠−3 1 1 1 = − 𝑒 −𝑡 + 𝑒 3𝑡 + 𝑒 −5𝑡 16 32 32 Question #20 + 1 32 𝐿−1 { }[using linearity property] 4{(𝑠−3)2 +22 } 1 𝑠+5 } 1 0  t  2  f (t ) = 2 2  t  5 4 t  5  use properties of Laplace transform Answer: Step 1 Convert the function into Heaviside's function. And then use properties of Laplace transform. The answer is given as follows: Step 2 1 0  t  2  f (t ) = 2 2  t  5 4 t  5  write function as heaviside function as 𝑓(𝑡) = 1 + (2 − 1)𝑢2 (𝑡) + (4 − 2)𝑢5 (𝑡) 𝑓(𝑡) = 1 + 𝑢2 (𝑡) + 2𝑢5 (𝑡) Now apply laplace transform and the formula 𝐿{𝑢𝑐 (𝑡)} = 𝑒 −𝑐𝑠 𝑠 ⇒ 𝐿{𝑓(𝑡)} = 𝐿{1} + 𝐿{𝑢2 (𝑡)} + 2𝐿{𝑢5 (𝑡)} 1 𝑒 −2𝑠 𝑠 𝑠 𝐹(𝑠) = + +2 𝑒 −5𝑠 𝑠